On a remark in Foundations of mechanics, 2nd Edition, by Abraham and Marsden

I don't know if this question is appropriate to this site, but I posted here without an answer, so I tried this alternative.

Given a $2$-form $\omega$ on a manifold $M$, let us denote by $N$ the kernel of $\omega$, i.e. $N:=\{u\in TM : \omega(u,\cdot)=0\}$. Their Proposition 5.1.2 shows that if $\omega$ has constant rank (and is closed) then $N$ is a tangent distribution on $M$ (and completely integrable).

In the following remark they say that the reader can easily prove the converse of the previous conclusion''. While I understand that $N$ is a tangent distribution if and only if $\omega$ has constant rank. Instead I think that, for $\omega$ of constant rank, $N$ can be completely integrable even if $\omega$ is not closed, (e.g. $\omega=e^z dx\wedge dy$).

Starting from this consideration I have asked myself a question:
Given a $\Omega\in\mathcal{A}^p(M)$, with $p>1$, whose rank is constant, let us define its kernel $N$ as above. Evidently $N$ is a tangent distribution on $M$, and I find it is completely integrable at least when there exists a $1$-form $\phi$ such that $d\Omega=\phi\wedge\Omega$. Clearly, if $\Omega$ is decomposable then the last condition is even necessary.

My question (edited after the comment of Willie Wong):

Is this last condition (the divisibility'' of $d\Omega$ by $\Omega$) necessary for the complete integrability of $N$ even when $\Omega$ is not decomposable? (Using Frobenius' Theorem I understand the case $p=1$, but what about the case $p>1$?.)

Any suggestion and\or counterexample are welcome.

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Presumably you know about Frobenius' theorem? en.wikipedia.org/wiki/Frobenius_theorem_(differential_topology) –  Willie Wong Aug 28 '11 at 21:45
@Willie Wong, thank for your suggestion, it has helped me to point out what is my doubt. I don't understand how to use Frobenius' Theorem to deduce, when degree(Ω)>1, that the complete integrability of kerΩ implies the divisibility of dΩ by Ω. What am I missing? –  Giuseppe Tortorella Aug 29 '11 at 11:27

The answer is 'no'. To see why, just take any nondegenerate $2$-form $\omega$ on, say, $\mathbb{R}^6$, that has the property that $d\omega$ is not a multiple of $\omega$. (This will be true for a generic such $2$-form.) The kernel of this $\omega$ is trivial, but now, you can just regard it as being defined on $\mathbb{R}^8$, say, by pullback. Then the kernel has (constant) positive dimension and is obviously integrable, but $d\omega$ is still not a multiple of $\omega$.
By the way, the theorem about integrability is true in much greater generality than for a single $p$-form: If $\mathcal{I}\subset\Omega^*(M)$ is a graded ideal that is closed under exterior derivative and its `kernel' $N$, which is defined as the set of vectors $v\in T_xM$ such that $\iota(v)\phi$ is in $\mathcal{I}_x$ for all $\phi\in \mathcal{I}_x$, has constant rank, then $N$ is integrable. This is a classic theorem of Cartan on 'Cauchy characteristics', and can, for example, be seen in Exterior Differential Systems by Bryant, et al.